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Viefleur [7K]
3 years ago
5

If 100 J of heat is added to a system so that the final temperature of the system is 400 K, what is the change in entropy of the

system? a)- 0.25 J/K b)- 2.5 J/K c)- 1 J/K d)- 4 J/K
Engineering
1 answer:
Vitek1552 [10]3 years ago
4 0

Answer:

0.25 J/K

Explanation:

Given data in given question

heat (Q) = 100 J

temperature (T) = 400 K

to find out

the change in entropy of the given system

Solution

we use the entropy change equation here i.e  

ΔS = ΔQ / T           ...................a

Now we put the value of heat (Q) and Temperature (T) in equation a

ΔS is the entropy change, Q is heat and T is the temperature,  

so that

ΔS = 100/400 J/K

ΔS = 0.25 J/K

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Answer:

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2 years ago
A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 55 MPa √m (50 ksi √in.). If, during
astra-53 [7]

Answer:

0.024 m = 24.07 mm

Explanation:

1) Notation

\sigma_c = tensile stress = 200 Mpa

K = plane strain fracture toughness= 55 Mpa\sqrt{m}

\lambda= length of a surface crack (Variable of interest)

2) Definition and Formulas

The Tensile strength is the ability of a material to withstand a pulling force. It is customarily measured in units (F/A), like the pressure. Is an important concept in engineering, especially in the fields of materials and structural engineering.

By definition we have the following formula for the tensile stress:

\sigma_c=\frac{K}{Y\sqrt{\pi\lambda}}   (1)

We are interested on the minimum length of a surface that will lead to a fracture, so we need to solve for \lambda

Multiplying both sides of equation (1) by Y\sqrt{\pi\lambda}

\sigma_c Y\sqrt{\pi\lambda}=K   (2)

Sequaring both sides of equation (2):

(\sigma_c Y\sqrt{\pi\lambda})^2=(K)^2  

\sigma^2_c Y^2 \pi\lambda=K^2   (3)

Dividing both sides by \sigma^2_c Y^2 \pi we got:

\lambda=\frac{1}{\pi}[\frac{K}{Y\sigma_c}]^2   (4)

Replacing the values into equation (4) we got:

\lambda=\frac{1}{\pi}[\frac{55 Mpa\sqrt{m}}{1.0(200Mpa)}]^2 =0.02407m

3) Final solution

So the minimum length of a surface crack that will lead to fracture, would be 24.07 mm or more.

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